1. ## Verifying Identities

(1+cscx/secx) - cotx = cosx

secx + tanx = (cosx/1-sinx)

(sinx/1-cosx) = (1+cosx/sinx)

2. Originally Posted by nichole23
(1+cscx/secx) - cotx = cosx
Hi nichole23,

Try posting these one at a time. I think you'd have a better chance of getting them answered. Just a suggestion.

$\frac{1+\csc x}{\sec x}-\cot x=\cos x$

$\dfrac{1+\dfrac{1}{\sin x}}{\dfrac{1}{\cos x}}-\dfrac{\cos x}{\sin x}=$

$\dfrac{\dfrac{\sin x+1}{\sin x}}{\dfrac{1}{\cos x}}-\dfrac{\cos x}{\sin x}=$

$\dfrac{\cos x(\sin x+1-1)}{\sin x}=$

$\dfrac{\cos x(\sin x)}{\sin x}=$

$\cos x$

3. Originally Posted by nichole23

secx + tanx = (cosx/1-sinx)
Hi again nichole23,

$\sec x+\tan x=\frac{\cos x}{1-\sin x}$

$\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{\cos x}{1-\sin x}$

$\frac{1+\sin x}{\cos x}=\frac{\cos x}{1-\sin x}$

Multiply left side by $\frac{1-\sin x}{1-\sin x}$

$\frac{1-\sin^2 x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$

$\frac{\cos^2 x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$

$\frac{\cos x}{1-\sin x}=\frac{\cos x}{1-\sin x}$

4. Originally Posted by nichole23

(sinx/1-cosx) = (1+cosx/sinx)
Hello again,

$\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}$

Multiply left side by $\frac{1+\cos x}{1+\cos x}$

$\frac{\sin x(1+\cos x)}{1-\cos^2 x}=\frac{1+\cos x}{\sin x}$

$\frac{\sin x(1+\cos x)}{\sin^2 x}=\frac{1+\cos x}{\sin x}$

$\frac{1+\cos x}{\sin x}=\frac{1+\cos x}{\sin x}$

5. Thank you so much!!!! I really appreciate it!